It would be great if you could explain me this base method plz.. and where I can apply these?
@Vinaysastra said:It would be great if you could explain me this base method plz.. and where I can apply these?
we can use only these digits 0,1,4,6,8,9
its like number system i base 6 where 0,1,2,3,4,5 are used...
its equivalent to:
0=0
1=1
4=2
6=3
8=4
9=5
so if we are asked to find 100th number inthe given case then
its nothing but finidng 100 in base 6 and replacing it wid the corresponding numbers that we have ...........
i think u can relate to it now...
@bodhi_vriksha said:A question from my side now:I write all natural numbers starting from one except the ones whose any digit is prime. First 10 numbers that I write are; 1, 4, 6, 8, 9, 10, 11, 14, 16, 18.What will be the 100th number that I write?Team BV
488
@Dexian said:214 ??
2 is prime.. 200 ka koi bhi number cannot be the answer
@saurav205 said:@ScareCrow28@Subhashdec2@bodhi_vriksha and others ..Next question :The last digit of the LCM of (3^2003 - 1) and (3^2003 + 1) is :8246
4?
used a weird method.
not sure.
both the numbers are consecutive even numbers. eg 4 and 6. .. they always have 2 has hcf.
product of 2 numbers = hcf * lcm
(3^2003 - 1)*(3^2003 + 1) = 2*lcm
lcm = (3^2003 - 1)*(3^2003 + 1)/2
= (3^4006 - 1)/2
3 has a cycle of 4. or the last digit will be..
= (9-1)/2
=4
used a weird method.
not sure.
both the numbers are consecutive even numbers. eg 4 and 6. .. they always have 2 has hcf.
product of 2 numbers = hcf * lcm
(3^2003 - 1)*(3^2003 + 1) = 2*lcm
lcm = (3^2003 - 1)*(3^2003 + 1)/2
= (3^4006 - 1)/2
3 has a cycle of 4. or the last digit will be..
= (9-1)/2
=4
@ChirpiBird said:4?used a weird method. not sure.both the numbers are consecutive even numbers. eg 4 and 6. .. they always have 2 has hcf.product of 2 numbers = hcf * lcm (3^2003 - 1)*(3^2003 + 1) = 2*lcmlcm = (3^2003 - 1)*(3^2003 + 1)/2= (3^4006 - 1)/23 has a cycle of 4. or the last digit will be.. = (9-1)/2=4
@ScareCrow28 said:488 ?? 1s --> 510s --> 640s --> 660s --> 680s--> 690s --> 65 + 6x = 100x = 15.xx And so on.. Till 479 there are 95 numbers..Hence 100th number = 488 ..
maine bhi yehi kia... didnt know the base thing.. 
@ChirpiBird said:maine bhi yehi kia... didnt know the base thing..
Haha! 
I knew but didn't use On the same boat, I suppose! 
I knew but didn't use On the same boat, I suppose! #Don't-use-base-boat
Amar, Bhavan, Chetan and Dinesh contributed a total of र240 to purchase a gift. Amar contributed half of the total contribution of the others. Bhavan contributed one-third of the total contribution of the others. Dinesh contributed र4 more than the contribution of Chetan. How much did Dinesh contribute (in र)?
@Faruq said:Amar, Bhavan, Chetan and Dinesh contributed a total of र240 to purchase a gift. Amar contributed half of the total contribution of the others. Bhavan contributed one-third of the total contribution of the others. Dinesh contributed र4 more than the contribution of Chetan. How much did Dinesh contribute (in र)?
52?
a + b + c + d = 240
a = 1/2(b + c + d)
b + c + d = 2a
a + 2a = 240
a = 80
b = 1/3(a + c + d)
a + c + d = 3b
b + 3b = 240
b = 60
c + d = 240 - 60 - 80
c + d = 100
d = 4 + c
d - 4 + d = 100
2d = 104
d = 52
@Faruq said:Amar, Bhavan, Chetan and Dinesh contributed a total of र240 to purchase a gift. Amar contributed half of the total contribution of the others. Bhavan contributed one-third of the total contribution of the others. Dinesh contributed र4 more than the contribution of Chetan. How much did Dinesh contribute (in र)?
52 ?
b + 3b = 240
=> b = 60
a + 2a = 240
=> a = 80
=> c + d = 100
=> 2c + 4 = 100
=> c = 48
Hence, d = 52 ..
Amar, Bhavan, Chetan and Dinesh contributed a total of र240 to purchase a gift. Amar contributed half of the total contribution of the others. Bhavan contributed one-third of the total contribution of the others. Dinesh contributed र4 more than the contribution of Chetan. How much did Dinesh contribute (in र)?
a+b+c+d=240
a=(b+c+d)/2
b=(a+c+d)/3
d=c+4
=> b = (a+2c+4)/3
=> a = ((a+2c+4/3) + 2c+4)/2 = (a+2c+4+6c+12)/6
=> 5a = 8c + 16
=> a = (8c+16)/5
=> b = ((8c+16)/5 + 2c + 4)/3 = (18c+26)/15
=> (8c+16)/5 + (18c+26)/15 + 2c = 236
=> 24c+48+18c+26+30c = 3540
=> 72c = 3466
=> c = 48.13
so d ~ 52 ??
lol fraction mein aya....matlab some error in calc....aese hi karlo bhai...office mein hun...srry...
If the six-digit number 15x67y is divisible by 56, then the value of x + y is
@ChirpiBird said:4?used a weird method. not sure.both the numbers are consecutive even numbers. eg 4 and 6. .. they always have 2 has hcf.product of 2 numbers = hcf * lcm (3^2003 - 1)*(3^2003 + 1) = 2*lcmlcm = (3^2003 - 1)*(3^2003 + 1)/2= (3^4006 - 1)/23 has a cycle of 4. or the last digit will be.. = (9-1)/2=4
Whats weird is this....standard procedure hai yeh toh....
City Central Multiplex employs 64 men to complete the construction of a 5 level parking lot. The men who work for 8 hours a day, manage to complete the construction of two levels in 9 days. If they are to complete the remaining 3 levels in 8 more days, how many hours a day should they be working?
@Faruq said:If the six-digit number 15x67y is divisible by 56, then the value of x + y is
x = 4, y = 2..so x + y = 6 ?
@Faruq said:City Central Multiplex employs 64 men to complete the construction of a 5 level parking lot. The men who work for 8 hours a day, manage to complete the construction of two levels in 9 days. If they are to complete the remaining 3 levels in 8 more days, how many hours a day should they be working?
13.5 ??
64*8*9*5/2 - 64*8*9 = 64*8*n
=> n = 13.5 ..
@Faruq said:City Central Multiplex employs 64 men to complete the construction of a 5 level parking lot. The men who work for 8 hours a day, manage to complete the construction of two levels in 9 days. If they are to complete the remaining 3 levels in 8 more days, how many hours a day should they be working?
13.5 ?
2/64*8*9 = 3/64*8*x = 13.5 hrs